vⁱ

and the action of the matrix A on vⁱ as

a^j_i v^j (assuming the summation convention).

Re: your para ‘If you define your matrix as A = aⁱ_j, then the upper index is indicating the row number and the lower the column number. For a vector, which has n rows but only one column, the index should go in the upstairs location for consistency v = v^j.’

I agree - this makes more sense.

a^j_i v_j

(assuming the summation convention) seems more straightforward to me. Employing upper and lower indices makes it much easier to keep a grip on summations.

This notation is common in differential geometry and hence relativity (and hence Penrose).

Depending on what I’m doing I usually write a_ij or aⁱ_j. For a lot of things the latter is more natural. (Especially when working with quantum doubles).

If you define your matrix as A = aⁱ_j, then the upper index is indicating the row number and the lower the column number. For a vector, which has n rows but only one column, the index should go in the upstairs location for consistency v = v^j. The transformation equation then becomes

v’ ⁱ = aⁱ_jv^j,

where we’ve followed Einstein’s summation convention of summing over dummy indices (provided one index is in the floor and the other index is in the ceiling).

So, what’s v_j I hear you ask ? Well, it’s an element of the dual space, a co-vector or 1-form. Skipping a lot of details (and precision), the two are related by the Kronecker delta which we use to change vectors into co-vectors and vice versa

v_i=δⁱ_jv^j.

Maybe you’ve had a sneaking suspicion all along about the transpose of a column vector :).

(Drinfeld’s quantum double combines a Hopf algebra H (actually H^Cop) with its dual H^* to create a new Hopf algebra which is always quasi-triangular.)

Tel

Hey, like you, I write the components of a matrix A as a_ij where i is the row and j is the column.

Now, in Penrose’s tome, he writes it as a_jⁱ. Do you know if this common physicists notation?

When I invented my own tensor notation, I wrote the action of the matrix A on a basis vector v_i as
a^j_iv_j (assuming the summation convention), but in Penrose’s scheme, it is
a_i^jv^j.

Cheers,
Sach

Hey, the definition of the tensor product of two matrices V and W is dependent on the basis of V \otimes W you choose, isn’t it? Ok, you’re implicitly choosing the basis in writing out V \otimes W. The way I’ve always dealt with it is just to be really clear about which basis I’m using. You know, I never actually learnt that “the definition of V \otimes W is…” I just worked it out myself!

Cheers,
Sach